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Standard IX

Mathematics

Area of a Triangle

ABCD is a par...

Question

ABCD is a parallelogram, G is the point on AB such that $A G = 2 G B$ . E is a point on DC such that $C E = 2 D E$ and F is a point of BC such that $B F = 2 F C$ . Prove that
$a r (δ E G B) = \frac{1}{6} a r (A B C D)$

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Solution

REF.Image.
$a r (E G B) = \frac{1}{2} \times x \times h$
$a r (A B C D) = 3 x \times h$
$\frac{a r (E G B)}{a r (A B C D)} = \frac{\frac{1}{2} x h}{3 x h} = \frac{1}{6}$
$\frac{a r (E G B)}{a r (A B C D)} = \frac{1}{6} \Rightarrow a r (E G B) = \frac{1}{6} a r (A B C D)$
Hence proved

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Similar questions

Q. ABCD is a parallelogram, G is the point on AB such that

A G = 2 G B

. E is a point on DC such that

C E = 2 D E

and F is a point of BC such that

B F = 2 F C

. Prove that

a r (△ E F C) = \frac{1}{2} a r (△ E B F)

Q. ABCD is a parallelogram, G is the point on AB such that

A G = 2 G B

. E is a point on DC such that

C E = 2 D E

and F is a point of BC such that

B F = 2 F C

.Then area of

△ E F G

\frac{5}{18}

of area of ABCD.

Q. ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(i) ar (ADEG) = ar (GBCE)

(ii) ar (Δ EGB) =

\frac{1}{6}

ar (ABCD)

(iii) ar (Δ EFC) =

\frac{1}{2}

ar (Δ EBF)

(iv) ar (Δ EBG) = ar (Δ EFC)

(v) Find what portion of the area of parallelogram is the area of Δ EFG

ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2 DE and F is the point of BC such that BF = 2 FC. Prove that:
$(i) a r (A D E G) = a r (G B C E) (i i) a r (△ E G B) - \frac{1}{6} a r (A B C D) (i i i) a r (△ E F C) = \frac{1}{2} a r (△ E B F) (i v) a r (△ E G B) = \frac{3}{2} a r (△ E F C)$
(v) Find what portion of the ara of parallelogram is the area of $△ E F G .$

A B C D

is a parallelogram.

E

is a point on

B A

such that

B E = 2 E A

and

F

is a point on

D C

such that

D F = 2 F C

. Prove that

A E C F

is a parallelogram whose area is one-third of the area of parallelogram

A B C D

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ABCD is a parallelogram, G is the point on AB such that AG=2GB. E is a point on DC such that CE=2DE and F is a point of BC such that BF=2FC. Prove that ar(δEGB)=16ar(ABCD)

REF.Image.ar(EGB)=12×x×har(ABCD)=3x×har(EGB)ar(ABCD)=12xh3xh=16ar(EGB)ar(ABCD)=16⇒ar(EGB)=16ar(ABCD)Hence proved

ABCD is a parallelogram, G is the point on AB such that $A G = 2 G B$ . E is a point on DC such that $C E = 2 D E$ and F is a point of BC such that $B F = 2 F C$ . Prove that
$a r (δ E G B) = \frac{1}{6} a r (A B C D)$

REF.Image.
$a r (E G B) = \frac{1}{2} \times x \times h$
$a r (A B C D) = 3 x \times h$
$\frac{a r (E G B)}{a r (A B C D)} = \frac{\frac{1}{2} x h}{3 x h} = \frac{1}{6}$
$\frac{a r (E G B)}{a r (A B C D)} = \frac{1}{6} \Rightarrow a r (E G B) = \frac{1}{6} a r (A B C D)$
Hence proved